We have the semi-ellipse curve $C$ defined by
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
where
$$x \le 0$$
Assuming clock-wise orientation (from $(0,-2)$ to $(0,2)$) of this curve,
find the unit tangent vector and the principal normal vector at the
point $P = (-3/\sqrt{2}, 2/\sqrt{2})$
Also find the curvature at that point.
I was able to solve this one but I just want to compare my answers.
Here are my answers:
$T = (3/\sqrt{13}, 2/\sqrt{13}, 0)$
$N = (2/\sqrt{13}, -3/\sqrt{13}, 0)$
$\kappa = (24 \cdot \sqrt{13} ) / 169$
The parametrization which I came up with and which I used was this one
$x = -3 \cos{t}$
$y = 2 \sin{t}$
$t \in [-\pi/2, \pi/2]$
When $t=\pi/4$ we get the point $P$ on the semi-ellipse.
Everything seems fine to me, except the curvature. Probably it's due a small calculation mistake. For a curve expressed in terms of a generical parametrization $r(t)=(x(t),y(t),z(t))$, we have that: $$\kappa(t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}$$ Your curve is parametrize by $r(t)=\left(-3\cos t,2\sin t,0\right)$, so that $r'(t)=(3\sin t,2\cos t,0)$ and $r''(t)=(3\cos t,-2\sin t,0)$. The cross product is: $$r'(t)\times r''(t)=\begin{vmatrix} \textbf i & \textbf j & \textbf k \\ 3\sin t & 2\cos t & 0 \\ 3\cos t & -2\sin t & 0 \end{vmatrix}=(0,0,-6\sin^2t-6\cos^2t)=(0,0,-6)$$ So: $$\kappa(t)=\frac{6}{(9\sin^2t+4\cos^2t)^\frac{3}{2}}=\frac{6}{(5\sin^2t+4)^\frac{3}{2}}$$ In the point $P=r\left(\frac{\pi}{4}\right)$ the curvature is: $$\kappa\left(\frac{\pi}{4}\right)=\frac{6}{\left(\frac{5}{2}+4\right)^\frac{3}{2}}=\frac{12\sqrt 2}{13\sqrt{13}}=\frac{12\sqrt{26}}{169}$$